Distance between two points

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two points. Planes through each point parallel to the coordinate planes make a rectangular box (a cuboid) with PQ as a diagonal.

In the box, the angle at corner A of triangle PAQ is right, so PQ² = PA² + AQ². AQ is itself the hypotenuse of the right triangle ANQ, so AQ² = AN² + NQ².

The three edges are the coordinate differences: PA = y₂ − y₁, AN = x₂ − x₁ and NQ = z₂ − z₁. Hence PQ² = (x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)².

PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]. From the origin, OQ = √(x² + y² + z²).

The order of the points does not matter because every difference is squared.

Worked: P(1, −3, 4) and Q(−4, 1, 2) give PQ² = 25 + 16 + 4 = 45, so PQ = 3√5 units.

A quick own check: (3, 4, 12) is √(9 + 16 + 144) = 13 from the origin; its foot (3, 4, 0) on the floor is 5 from O, and 5, 12, 13 is the vertical right triangle.